we can obtain value \(\tan 15\) from \[\tan(45-30)=\frac{\tan45-\tan30}{1+\tan45\tan30} = \frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}\]
now, to obtain \(\tan7.5\) we use
\[\tan15=\frac{2\tan7.5}{1-\tan^27.5}\] which can solve by Quadratic formula